At an axial load of 22 kN, a 45-mm-wide × 15-mm-thick polyimide polymer bar elongates 3.0 mm while the bar width contracts 0.25 mm. The bar is 200 mm long. At the 22-kN load, the stress in the polymer bar is less than its proportional limit. Determine:
(a) the modulus of elasticity.
(b) Poisson’s ratio.
(c) the change in the bar thickness.

Question

Accepted Answer

(a) The bar cross-sectional area is

[latex]A=(15 mm )(45 mm )=675 mm ^{2}[/latex]

and thus, the normal stress corresponding to the 22-kN axial load is

[latex]\sigma=\frac{(22 kN )(1,000 N / kN )}{675 mm ^{2}}=32.592593 MPa[/latex]

The longitudinal strain in the bar is

[latex]\varepsilon_{ ext {long }}=\frac{\delta}{L}=\frac{3.0 mm }{200 mm }=0.0150 mm / mm[/latex]

The modulus of elasticity is therefore

[latex]E=\frac{\sigma}{\varepsilon_{ ext {long }}}=\frac{32.592593 MPa }{0.0150 mm / mm }=2,173 MPa =2.17 GPa[/latex]

(b) The longitudinal strain in the bar was calculated previously as

[latex]\varepsilon_{ ext {long }}=0.0150 mm / mm[/latex]

The lateral strain can be determined from the reduction of the bar width:

[latex]\varepsilon_{ lat }=\frac{\Delta ext { width }}{ ext { width }}=\frac{-0.25 mm }{45 mm }=-0.005556 mm / mm[/latex]

Poisson’s ratio for this specimen is therefore

[latex]v=-\frac{\varepsilon_{ ext {lat }}}{\varepsilon_{ ext {long }}}=-\frac{-0.005556 mm / mm }{0.0150 mm / mm }=0.370370=0.370[/latex]

(c) The change in bar thickness can be found from the lateral strain:

[latex]\Delta ext { thickness }=\varepsilon_{ lat }( ext { thickness })=(-0.005556 mm / mm )(15 mm )=-0.0833 mm[/latex]

Question 3.3: At an axial load of 22 kN, a 45-mm-wide × 15-mm-thick polyim...

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